C strings confusion

Question

I\'m learning C right now and got a bit confused with character arrays - strings.

char name[15]=\"Fortran\";

No problem with this - its an

Answer 1

In C a string is actually just an array of characters, as you can see by the definition. However, superficially, any array is just a pointer to its first element, see below for the subtle intricacies. There is no range checking in C, the range you supply in the variable declaration has only meaning for the memory allocation for the variable.

a[x] is the same as *(a + x), i.e. dereference of the pointer a incremented by x.

if you used the following:

char foo[] = "foobar";
char bar = *foo;

bar will be set to 'f'

To stave of confusion and avoid misleading people, some extra words on the more intricate difference between pointers and arrays, thanks avakar:

In some cases a pointer is actually semantically different from an array, a (non-exhaustive) list of examples:

//sizeof
sizeof(char*) != sizeof(char[10])

//lvalues
char foo[] = "foobar";
char bar[] = "baz";
char* p;
foo = bar; // compile error, array is not an lvalue
p = bar; //just fine p now points to the array contents of bar

// multidimensional arrays
int baz[2][2];
int* q = baz; //compile error, multidimensional arrays can not decay into pointer
int* r = baz[0]; //just fine, r now points to the first element of the first "row" of baz
int x = baz[1][1];
int y = r[1][1]; //compile error, don't know dimensions of array, so subscripting is not possible
int z = r[1]: //just fine, z now holds the second element of the first "row" of baz

And finally a fun bit of trivia; since a[x] is equivalent to *(a + x) you can actually use e.g. '3[a]' to access the fourth element of array a. I.e. the following is perfectly legal code, and will print 'b' the fourth character of string foo.

#include 

int main(int argc, char** argv) {
  char foo[] = "foobar";

  printf("%c\n", 3[foo]);

  return 0;
}